A formula for the area in terms of local parametrisation Let $X(s,t)$ be a local parametrisation then
$\mid \mid X_{v} \times X_{u}\mid \mid = \sqrt{EG-F^2}$
where $E=<X_{u},X_{u}>$, $F=<X_{u},X_{v}>$ and $G=<X_{v},X_{v}>$, the components of the first fundamental form.
Does anyone know a proof of this?
 A: This is pretty easy to see if we recall that
$\Vert X_u \times X_v \Vert = \Vert X_u \Vert \Vert X_v \Vert \vert \sin \theta \vert, \tag 1$
where $\theta$ is the angle 'twixt the vectors $X_u$ and $V_v$; then
$\Vert X_u \times X_v \Vert^2 = \Vert X_u \Vert^2 \Vert X_v \Vert^2 \vert \sin \theta \vert^2; \tag 2$
likewise $F$ may be expressed in terms of $\cos \theta$:
$F = \langle X_u, X_v \rangle = \Vert X_u \Vert \Vert X_v \Vert \cos \theta, \tag 3$
whence
$F^2 = \Vert X_u \Vert^2 \Vert X_v \Vert^2 \vert \cos \theta \vert^2; \tag 4$
then
$\Vert X_u \times X_v \Vert^2 + F^2 = \Vert X_u \Vert^2 \Vert X_v \Vert^2 \vert \sin \theta \vert^2 + \Vert X_u \Vert^2 \Vert X_v \Vert^2 \vert \cos \theta \vert^2$
$= \Vert X_u \Vert^2 \Vert X_v \Vert^2(\vert \sin \theta \vert^2 + \vert \cos \theta \vert^2 ) = \Vert X_u \Vert^2 \Vert X_v \Vert^2; \tag 5$
also,
$E = \langle X_u, X_u \rangle = \Vert X_u \Vert^2; \tag 6$
$G = \langle X_v, X_u \rangle = \Vert X_v \Vert^2; \tag 7$
we fold (6) and (7) into (5), yielding
$\Vert X_u \times X_v \Vert^2 + F^2 = EG, \tag 8$
that is,
$\Vert X_u \times X_v \Vert^2 = EG - F^2, \tag 8$
and finally
$\Vert X_u \times X_v \Vert = \sqrt{EG - F^2}, \tag 9$
$OE\Delta$.
For more on formulas (1) and (3), see this Wikipedia page.
